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t the correct answerAnalyse the code and choose the right option.#include <stdio.h>int main(){  extern int A;  printf("%d",A);  return 0;}A=550;

Question

Analyse the code and choose the right option.

#include &lt;stdio.h&gt;
int main(){
    extern int A;
    printf(&quot;%d&quot;, A);
    return 0;
}

A = 550;
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Solution

The code you provided is written in C language. It's trying to print the value of an external integer variable 'A'. However, the declaration of 'A' is outside the main function and after it, which is not correct in C.

In C, if a variable is declared as extern, it means it has been defined elsewhere, not within the same block where it is declared. The extern keyword is used to extend the visibility of variables/functions.

Since the variable 'A' is not properly defined and initialized before the main function, the program will throw an error.

So, the correct answer is: The code will throw an error because the external variable 'A' is not defined before the main function.

This problem has been solved

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